Question

The x - z plane is the boundary between two transparent media. Medium 1 with $z\ge 0$ has a refractive index $\sqrt{2}$ and medium 2 with $z\le 0$ has a refractive index $\sqrt{3}$. A ray of light in medium 1 given by the vector $A=6\sqrt{3}\hat{i}+8\sqrt{3}\hat{j}-10\hat{k}$ is incident on the plane of separation. The angle between vector A and the positive z - direction is

• A. ${90}^{\circ }$
• B. ${120}^{\circ }$
• C. ${135}^{\circ }$
• D. ${150}^{\circ }$

Answer 1

The correct option is B

${120}^{\circ }$

Refer to Fig. Ray PQ in x-z plane travelling in medium 1 $(z\ge 0)$ at an angle i on the boundary (z = 0 plane) and is refracted along QR in medium 2 $(z\le 0)$ at an angle of reflection r.

Let $\theta$ be the angle between vector A and the positive z - direction. Since $\hat{k}$ is the unit vector along the positive z - axis, we have

$cos\theta =\frac{A.\hat{k}}{|A|}$

$\frac{(6\sqrt{3}\hat{i}+8\sqrt{3}\hat{j}-10\hat{k}).\left(\hat{k}\right)}{{\left[\right(6\sqrt{3}{)}^2+(8\sqrt{3}{)}^2+(-10{)}^2.]}^{1/2}}$

$=\frac{-10}{20}=-\frac{1}{2}$

Which gives $\theta ={120}^{\circ }$ .